What is the probability of finding the second qubit as $0$ in the state $|psirangle=frac1{sqrt2}|00rangle+frac12|10rangle-frac12|11rangle $?

Question

Assuming two qubits start in the state:
$|psirangle = frac{1}{sqrt 2}|00rangle +  frac{1}{2}|10rangle- frac{1}{2}|11rangle $
What is the probability of measuring the second qubit as 0? And what is the new state of the system after measuring the first qubit as 1?
I know that for a single qubit state that the probability amplitudes are the coefficient squared. In a two qubit system are the probabilities distributed to the individual states? I.e. from this example does the each zero state in the state: $|00⟩$ have a 50% chance?  And I don't really understand the second question, any suggestion on where to review or study?

KAJ226 · Accepted Answer

If we have the state $|psi rangle = dfrac{1}{sqrt{2}}|00rangle + dfrac{1}{2}|10rangle - dfrac{1}{2}|11rangle$ then the probability of the second qubit being in the state $|0rangle$ is the probability of the state $|psi rangle$ having $|0rangle$ on the second qubit. In this case, it is from the states $|00rangle$ and $|10rangle$. So The probability of measuring the second qubit in the state $|0rangle$ is $bigg| dfrac{1}{sqrt{2}} bigg|^2 + bigg| dfrac{1}{2} bigg|^2 = dfrac{3}{4} $.
You can also work this out more explicitly as well. That is, we have
$$
|psi rangle = begin{pmatrix} 1/sqrt{2}    0  1/2  -1/2 end{pmatrix}
$$
We are looking for the probability that the second qubit is in the state $|0rangle$ so the projective measurement $M$ is
$$ 
M = I otimes |0rangle langle 0 | = begin{pmatrix}  
     1   &  0   &  0  &   0
     0  &   0   &  0   &  0
     0   &  0   &  1 &    0
     0    & 0   &  0   &  0
 end{pmatrix}
$$
and so according to Born's rule we have that the probability to measure the second qubit in the state $|0rangle$ is
$$
 langle psi | M | psi rangle =  begin{bmatrix} 1/sqrt{2} & 0 & 1/2  &-1/2 end{bmatrix} 
begin{bmatrix}  
     1   &  0   &  0  &   0
     0  &   0   &  0   &  0
     0   &  0   &  1 &    0
     0    & 0   &  0   &  0
 end{bmatrix} begin{bmatrix} 1/sqrt{2}    0  1/2  -1/2 end{bmatrix} = dfrac{1}{2} + dfrac{1}{4} = dfrac{3}{4}
$$

Also, the state post measurement is $|psi_{post} rangle = dfrac{M|psirangle}{sqrt{3/4}}$.
You can extend this to the case where the first qubit is mesured in the state $|1rangle$ too. In this case, the projective measurement $M = |1rangle langle 1| otimes I$

What is the probability of finding the second qubit as $0$ in the state $|psirangle=frac1{sqrt2}|00rangle+frac12|10rangle-frac12|11rangle $?

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